Problem 34 - University Physics with Modern Physics

Question

A uniform solid cylinder of mass \(M=5.00 \mathrm{~kg}\) is rolling without slipping along a horizontal surface. The velocity of its center of mass is \(30.0 \mathrm{~m} / \mathrm{s}\). Calculate its energy.

Step-by-Step Solution

Answer

Answer: The total energy of the uniform solid cylinder is 6,750 J.

Step 1: Write down the given values

Mass of the cylinder, \(M = 5.00 \, kg\). Velocity of the center of mass, \(v = 30.0 \, m/s\).

Step 2: Relate linear velocity to angular velocity

Since the cylinder is rolling without slipping, its linear velocity \(v\) is related to its angular velocity \(\omega\) by: \(v = R\omega\) We will need this relation later.

Step 3: Find the moment of inertia of the cylinder

The moment of inertia of a uniform solid cylinder rotating about its axis is given by the formula: \(I = \frac{1}{2}mR^2\) Note that we have not been given the radius of the cylinder, so we will leave the moment of inertia in terms of \(R\).

Step 4: Calculate the rotational kinetic energy

The rotational kinetic energy is given by the formula: \(KE_{rot} = \frac{1}{2}I\omega^2\) Replacing \(I\) with the expression found in step 3, we get: \(KE_{rot} = \frac{1}{2} \times \frac{1}{2}mR^2 \times \omega^2\) Using the relation from step 2, \(\omega = \frac{v}{R}\), we can replace \(\omega\) in the expression for \(KE_{rot}\): \(KE_{rot} = \frac{1}{2} \times \frac{1}{2}mR^2 \times \left(\frac{v^2}{R^2}\right)\) Simplifying, we get: \(KE_{rot} = \frac{1}{4}m v^2\)

Step 5: Calculate the translational kinetic energy

The translational kinetic energy is given by the formula: \(KE_{trans} = \frac{1}{2}mv^2\)

Step 6: Find the total energy

Now we can find the total energy by adding the translational and rotational kinetic energies: \(E_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{4}m v^2\) Combining terms, we get: \(E_{total} = \frac{3}{4}mv^2\)

Step 7: Calculate the energy using the given values

Finally, substitute the given values for mass and velocity into the expression for \(E_{total}\): \(E_{total} = \frac{3}{4}(5.00\, kg)(30.0\, m/s)^2\) After performing the calculations, we get: \(E_{total} = 6,750\, J\) The total energy of the uniform solid cylinder rolling without slipping along a horizontal surface is \(6,750\, J\).